Electrons distribution when a conductor is placed in an electric field

Question

(This is releated to the question of how electric field inside a conductor is zero, but Im trying to focus on a specific part of that expalnation in this question, I hope you understand).

Im considering the simplest case of a large sheet of small thickness placed in a uniform electric field(figure 1.0). (I have shown its cross section in figure 1.1)

Usually in textbooks its shown that the sheet acquires some negative charge on left surface and some positive charge on right surface.

But what's actually happening inside the sheet with rest of the electrons?

Here's what I think is happening-

"All the electrons inside the sheet are pulled towards the right edge by the electric field. The electrons which are nearest to the left edge get closer together and there is an increase in density at that edge. But, rest of the electrons are also pulled towards the left edge but they are pulled less and less as we go towards the right edge, hence creating an exponentially (or some other mathematical function) decreasing distribution of density of electrons as we go from left edge to the right edge(As shown in figure 2.0).

How correct am I in my understanding or something else is happening?

Answer 1

Yes that is roughly the picture. However, you must remember that for neutrality, there is a background of positive charge. Therefore a lack of electrons translates into an excess of charge. This is why it’s best to reason in terms of difference with the baseline density which will more easily translate in charge density (this is usually what you are more interested in). Also, from symmetry consideration, you would expect the distribution to present central symmetry, so it would rather be a hyperbolic sine, instead of just an exponential. Finally, the location of the electrons is not so important in most simple models.

The idea is that at the beginning the electric field brings the electrons to the left. This generates a pileup of negative charge at the left and a pileup of positive charge at the right. This creates an opposite field in the conductor, attenuating the effect of the original field.

What is not obvious is that the system relaxes to the equilibrium with no overshoot, you’ll need to do the math. Things may be clearer with equations.

I’ll restrict to the 1D case by lateral symmetry. Say your conductor has resistivity $\rho$ and the external electric field is $E_0$ parallel to the conductor.

Initially, there is only the external electric field which generates a current $$j=\rho^{-1} E_0$$ in the conductor. Since outside there is no current, from conservation of charge, this generates a accumulation of surface charge $\sigma_\pm$ at either end: $$ \dot \sigma_\pm=\pm j $$ These new charges in turn generate an electric field: $$ E= \mp\frac{\sigma_\pm}{\epsilon_0} $$ This modifies the original current where the field is now $E+E_0$. This gives a closed loop of equations, and in particular, you can deduce an ODE on the charge: $$ \dot \sigma_\pm+\frac{1}{\rho\epsilon_0}\sigma_\pm= \pm\frac{1}{\rho}E_0 $$ In particular, in this simple model, the electron density is flat in the conductor with two opposing Dirac peaks at either end for the surface charge. I also obtain the relaxation of charge distribution to the equilibrium value: $$ \sigma_\pm= \epsilon_0 E_0 $$ Actually, this is merely the continuum version of a RC circuit if you think about it.

Hope this helps.